Question: Simplify and expand the following expression: $ \dfrac{4z}{3z - 1}-\dfrac{4z + 4}{4z + 4} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3z - 1)(4z + 4)$ Multiply the first term by $\dfrac{4z + 4}{4z + 4}$ $ \begin{align*} \dfrac{4z}{3z - 1} \times \dfrac{4z + 4}{4z + 4} & = \dfrac{(4z)(4z + 4)}{(3z - 1)(4z + 4)} \\ & = \dfrac{16z^2 + 16z}{(3z - 1)(4z + 4)}\end{align*} $ Multiply the second term by $\dfrac{3z - 1}{3z - 1}$ $ \begin{align*} \dfrac{4z + 4}{4z + 4} \times \dfrac{3z - 1}{3z - 1} & = \dfrac{(4z + 4)(3z - 1)}{(4z + 4)(3z - 1)} \\ & = \dfrac{12z^2 + 8z - 4}{(4z + 4)(3z - 1)}\end{align*} $ Now we have: $ = \dfrac{16z^2 + 16z}{(3z - 1)(4z + 4)} - \dfrac{12z^2 + 8z - 4}{(4z + 4)(3z - 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{16z^2 + 16z - (12z^2 + 8z - 4)}{(3z - 1)(4z + 4)} $ $ = \dfrac{16z^2 + 16z - 12z^2 - 8z + 4}{(3z - 1)(4z + 4)} $ $ = \dfrac{4z^2 + 8z + 4}{(3z - 1)(4z + 4)}$ Expand the denominator: $ = \dfrac{4z^2 + 8z + 4}{12z^2 + 8z - 4}$ Simplify: $ = \dfrac{z^2 + 2z + 1}{3z^2 + 2z - 1}$